2 - Conic Sections-V2010

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The Conics

Conics, an abbreviation for conic sections, are

curves that result from the intersection of a

right circular cone and a plane.

Depending on the angle of the plane relative to

the cone, the intersection can be a parabola, anellipse, a circle, or a hyperbola.

Definition

Conics

The general form of a second degree equation in 2

variables x, y is given by

ax2 +bxy + cy2 + dx +ey + f = 0

where a, b, c, d, e, and f are constants and at least one

of a, b, and c is nonzero .

The shapes of the corresponding graphs of the

equation depend on the values of these constants.

General Form

Consider the second-degree equation of the form

ax2 +bxy + cy2 + dx +ey + f = 0

where a, b, c, d, e, and f are constants and at least oneof a, b, and c is nonzero .

The graph is

an ellipse if b2 ± 4ac < 0, b 0, a c

a circle if b2 ± 4ac < 0, b = 0, a = c

a parabola if b2 ± 4ac = 0

a hyperbola if b2 ± 4ac > 0

Identification of a Conic

Conics

parabola

b2 ± 4ac = 0

hyperbola

b2 ± 4ac > 0

ellipse

b2 ± 4ac < 0, b 0, a c

circleb2 ± 4ac < 0, b = 0, a = c

Identify the conic section.

1. x2+ y2 + 4x + 12y + 36 = 0

Solution:a = 1 , b = 0, c = 1 , d = 4, e = 12, and f = 36

b2 ± 4ac = (0)2 ± 4( 1 )( 1 ) = ± 4 < 0

a = c and b = 0

circle

Example

Identify the conic section.

2. 2x2 ± 4xy + 8y2 + 7 = 0

Solution:a = 2, b = ± 4, c = 8, d = 0, e = 0, and f = 7

b2 ± 4ac = (± 4 )2 ± 4(2)(8) = ± 48 < 0

a c and b 0

ellipse

Example

Determine the type of conic section that each equation

will produce.

1. x2+ y2 + 4x + 12y + 36 = 0

2. 2x2 ± 4xy + 8y2 + 7 = 03. x2 ± 8x ± 6y ± 8 = 0

4. 3 x2 + 6xy + 5y2 ± x + y = 0

5. 2 x2 + xy ± y2 + x ± y = 0

6. x2+ y2 ± 10x + 4y ± 7 = 07. x2+ y2 ± 10x ± 24y + 25 = 0

8. 2 x2 ± y2 + 4xy ± 2x + 3y ± 6 = 0

9. x2 + 4x +16y + 4 = 0

Example

1. circle

2. ellipse3. parabola

4. ellipse

5. hyperbola

6. circle7. circle

8. hyperbola

9. parabola

Parabola

Definition of a Parabola

A parabola is the set of all points in a plane

that are equidistant from a fixed line (the

directrix) and a fixed point (the focus) that is

not on the line.

Directrix

Parabola

Vertex

Axis of

Symmetry

Directrix x = - p

y 2 = 4 p x

Vertex

Focus ( p, 0)

Directrix y = - p

x 2 = 4 p y

Vertex

Focus ( p, 0) y

Standard Forms of the Parabola

The standard form of the equation of a parabola withvertex at the origin is

y2 = 4 p x or x 2 = 4 p y.

the focus is on the x -axis,which is the axis of symmetry.

the focus is on the y-axis,

which is the axis of symmetry.

Parabolas

Directrix

Parabola

Standard form of the equation

of a parabola with vertex (0,0)

EquationEquation FocusFocus DirectrixDirectrix AxisAxis

xx22=4py=4py (0,p)(0,p) y =y = --pp

yy22=4px=4px (p,0)(p,0) x =x = --pp

Find the focus and directrix of each parabola. Then graph

the parabola.

1. x2 = 16y 2. y2 = ± 12x 3. 2x2 + 5y = 0

Solution:

1. x2 = 16y

Vertex : (0, 0)

Focus: x2 = 16y

x2 = 4(4)y

(0,p) = (0,4)

Directrix: y = ±4

Example

-10 -5 5 10

y = -4

(-8, 4) (8, 4)

focus (0,4)

vertex (0,0)

Find the focus and directrix and then graph the parabola.

2. y2 = ± 12x

Solution:

2. y2 = ± 12x

Vertex : (0, 0)

Focus: y2 = ± 12x

y2 = 4(±3)x

p = ±3 (p,0) = (±3 ,0)

Directrix: x = 3

Example

directr ix x = 3

(-3, -6)

(-3, 6)

focus (-3, 0) ver tex (0,0)

Find the focus and directrix and then graph the parabola.3. 2x2 + 5y = 0

Solution:

3. 2x2

+ 5y = 0 x2 = ± 5/2 y

Vertex : (0, 0)

Focus: x2 = ±5/2 y

4p = ±5/2 p = (±5/2)( 1 /4)

(0,p) = (0, ±5/8)

Directrix: y = 5/8

Example

-10 -5 5 10

directr ix y = 5/8

focus (0, -5/8)

vertex (0, 0)

Standard Form of the Parabola with

Vertex at (h, k)

EquationEquation FocusFocus DirectrixDirectrix AxisAxis

(x(x--h)h)22=4p(y=4p(y--k)k) ((h,k+ph,k+p)) y =ky =k--pp

(y(y--k)k)22=4p(x=4p(x--h)h) ((h+p,kh+p,k)) x =hx =h--pp

For each of the following, find the vertex, focus, directrix,and axis of symmetry. Then graph the parabola.

1. (x ± 3 )2 = 10 (y + 2) 2. y2 ± 12x +48 = 0

Solution:

1. (x ± 3 )2

= 10 (y + 2)vertex: (h, k) = (3, ±2)

focus: (h, k + p) = ?

4p = 10

p = 5/2 (3, ½)

directrix: y = k ± p = ± 9/2

Axis of symmetry: x = 3

Example

1. (x ± 3 )2

= 10 (y + 2)

Example

- 6 -4 - 2 2 4 6 8 10

directr ix y = -9/2

focus (3, 1/2)

vertex (3, -2)

For each of the following, find the vertex, focus, directrix,and axis of symmetry. Then graph the parabola.

1. (x ± 3 )2 = 10 (y + 2) 2. y2 ± 12x +48 = 0

Solution:

± 12x +48 = 0 y2 = 12x ± 48

(y ± 0)2 = 12(x ± 4)

vertex: (h, k) = (4, 0)

focus: (h + p, k) = ? 4p = 12 p = 3 (7, 0)

directrix: x = h ± p = 4 ± 3 = 1

Axis of symmetry: y = 0

Example

± 12x +48 = 0(y ± 0)2 = 12(x ± 4)

Example8

5 10 15

di rectr ix x = 1

focus (7, 0)

vertex (4, 0)

Find an equation of the parabola with vertex

at the origin and focus (-2, 0).

Vertex: (0, 0); Focus: (-2, 0) = ( p, 0)

x y )2(42!

Example

px y 42!

x y 82!

Find an equation of the parabola with focus

at ( ± 10, ± 2) and vertex at (3, ± 2).

Vertex: (3, ± 2);

Focus:( ± 10, ± 2)

Example

h x pk y !

)3(422

! x p y

-10 -8 -6 -4 -2 2 4

vertex (3, -2)focus (-10, -2)

Find an equation of the parabola with focus

at ( ± 10, ± 2) and vertex at (3, ± 2).

Example

h x pk y !

01524522

! y x y

! x p ySince focus ( ± 10, ± 2), then F(h + p, k) h + p = ± 10

3 + p = ± 10

p = ± 13

)3)(13(422

Ellipse

Definition of an Ellipse

An ellipse is the set of all points in a plane such that the

sum of the distances of any point on the ellipse to two

other fixed points is a positive constant. These two fixed

points are called the foci.

An ellipse is the set of all points in a plane such that the

sum of the distances of any point on the ellipse to two

other fixed points is a positive constant. These two fixed

points are called the foci.

The midpoint of the segment connecting the foci is

the center of the ellipse.

The intersection of the ellipse and the line joining

the foci are the vertices of the ellipse.

The segment between the 2 vertices is called the

major axis of the ellipse.

The segment perpendicular to the major axis and

intercepted by the ellipse is called the minor axisof the ellipse.

(0, b)

(0, -b)

(0, a)

(0, -a)

(0, c)

(0, -c)

(0, 0) (a, 0) (0, 0)(-a, 0)

(c, 0)(-c, 0)

(b, 0)(-b, 0)

22221 xyab! 22221 xyba!

Standard Forms of the Equations of an Ellipse

The standard form of the equation of an ellipse with center at the origin, and major and minor axes of lengths 2aand 2b (where a and b are positive, and a2 > b2) is

(0, b)

(0, -b)

(0, a)

(0, -a)

(0, c)

(0, -c)

(0, 0) (a, 0) (0, 0)(-a, 0)

(c, 0)(-c, 0)

(b, 0)(-b, 0)

22221 xyab! 22221 xyba!

Standard Forms of the Equations of an Ellipse

The vertices are on the major axis, a units form thecenter. The foci are on the major axis, c units form thecenter. For both equations, c2 = a2 ± b2.

Find the vertices, foci, endpoints of the minor

axis, and lengths of the major and minor axes

of each ellipse. Then graph the ellipse.

Example

4001625.)1 22! y x 225259.)2 22

Find the vertices, foci, endpoints of the minor axis, and lengths of

the major and minor axes and then graph.

Example

4001625.)1 22! y x

The equation is the standard form of an ellipse¶s equationwith a2 = 25 and b2 = 16.

Because the denominator of the y2 term is greater than thedenominator of the x2 term, the major axis is vertical.

Find the vertices, foci, endpoints of the minor axis, and lengths of the major and minor axes and then graph.

Example

vertices are (0, -a) and (0, a) the vertices are (0, -5) and (0, 5).

endpoints of the horizontalminor axis are (-b, 0) and

(b, 0)

b2 = 16, b = 4 (-4, 0) and (4,0).

foci located at (0, -c) and (0,c)

c2 = a2 ± b2

a2 =25 and b2 = 16

c2 = a2 ± b2 = 25 ± 16 = 9.

(0, -3) and (0, 3)

(0, 5)

(0, -5)

(0, 3)

(0, -3)

(0, 0) (4, 0)(-4, 0)

Example

10)5(22 !!a 8)4(22 !!b

length of the major axis length of the minor axis

Find the vertices, foci, endpoints of the minor axis, and lengths of the major and minor axes and then graph.

Example

The equation is the standard form of an ellipse¶s equation witha2 = 25 and b2 = 9.

Because the denominator of the x2 term is greater than thedenominator of the y2 term, the major axis is horizontal.

122525

225259.)2 22! y x

Find the vertices, foci, endpoints of the minor axis, andlengths of the major and minor axes and then graph.

Example

vertices are (-a, 0) and (a,0) the vertices are (-5,0) and (5, 0).

endpoints of the verticalminor axis are (0, -b) and

(0, b)

b2 = 9, b = 3 (0, -3) and (0, 3).

foci located at (-c, 0) and(c, 0)

c2 = a2 ± b2,

a2 =25, b2 = 9,

c2 = a2 ± b2 = 25 ± 9 = 16

(4, 0) and (-4, 0).

F 2=(4, 0) F

1=(-4, 0)

V 2=(5, 0)V

(0, 3)

(0, -3)

= (-5, 0)

Example

10)5(22 !!a 6)3(22 !!b

length of the major axis length of the minor axis

Standard Forms of Equations of Ellipses Centered at (h,k)

Find the vertices, foci, endpoints of the minor axis, and

lengths of the major and minor axes of each ellipse. Then

graph the ellipse.

Example

0144722494.)1 22! x y x y

Example

0144722494.)122

! x y x y

4y2 + 9x2 ± 24y ± 72x + 144 = 0

4y2 ± 24y + 9x2 ± 72x = ± 144

4(y2 ± 6y) + 9(x2 ± 8x) = ± 144

4(y2 ± 6y + 9) + 9(x2 ± 8x + 16) = ± 144 + 36 + 144

4(y ± 3)2 + 9(x ± 4)2 = 36 1

Example

0144722494.)122

! x y x y

center: (4, 3)

Vertices: (4, 6) and (4, 0)

Endpoints of the minor axis: (2,3) and (6,3)

53,4 53,4

Example

center: ( 1 , 3)Vertices: ( 1 , 16) and ( 1 , ± 10)

Endpoints of the minor axis: ( 13,3) and (± 11 ,3)

Foci: ( 1 , 8) and ( 1 , ± 2)

1. 2x2 ± 4xy + 8y2 + 7 = 0

2. 3 x2 + 6xy + 5y2 ± x + y = 0

3. y2 ± 2xy +2 x2 ± 5x = 0

Example

Find the vertices, foci, endpoints of the minor axis,

and lengths of the major and minor axes of each

ellipse. Then graph the ellipse.

2 - Conic Sections-V2010

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